Procedure:
[1] It’s ideal to start with a commutator in which two of its 1x2x2 blocks are already in the same slice and are adjacent to each other (or if not, very close to it).
[2] Do moves to bring the third 1x2x2 block into the same slice as the other two (if none of the 1x2x2 blocks are in the same slice, then you need to move two of them into the same slice as one of them). If you desire to have shorter algorithms, you need to simultaneously do at least a portion of step [3] here.
[3] Move appropriate center pieces into the only block in slice r which was not originally involved in the 3-cycle as well as the appropriate wing edge there too (if you want to create a higher cycle, you need to substitute a wing edge for one or more of the three blocks affected by the 3-cycle).
[4] Add quarter turn and reverse conjugates.
Tip: It’s good to make the first moves of the conjugation the inverse of the last moves of the three 1x2x2 block cycler in order to have the most move cancellations.
Example
Assuming that green is the color of the front face, white is the color of the top face, and your cube has the Rubik’s color scheme,
[1] Start out with the commutator [Fw', Uw r' Uw'] =
Fw' Uw r' Uw' Fw Uw r Uw'.
By following the “tip,” our first move should be, without the shadow of a doubt, Uw.
Fw' Uw r' Uw' Fw Uw r Uw'
Uw
It works out well for us because it joins two of the 1x2x2 blocks together in the same slice (slice r, in this case).
[2] We have a change to follow the “tip” again by choosing to execute r2 next because it will partially cancel with the second to last move of the commutator, r, and it will make way to place in the third block affected (the red and white one).
Fw' Uw r' Uw' Fw Uw r Uw'
Uw r2
Before I go any further, we need to visualize in which direction the quarter turn should move in. We can figure this out by asking ourselves, “In which direction will solve back half of slice r?” In fact, let's temporally delete the last move, r2, we just added.
Fw' Uw r' Uw' Fw Uw r Uw'
Uw
By looking at the cube, shouldn’t it be r'? Yep. REASON: the green and yellow 1x2x2 block in the front top needs to move to the front bottom.
[3] Now adding Fw' so that the red and white 1x2x2 block is placed in slice r.
Fw' Uw r' Uw' Fw Uw r Uw'
Uw r2 Fw'
[4] Since we know that the direction of the quarter turn is r', let’s go ahead and at that and undo the setup moves we added so far so that we can see which two wing edges swap with each other. This way, we can see which wing edge we can substitute with another to make an alternate 2-cycle but yet swapping the same center pieces.
Fw' Uw r' Uw' Fw Uw r Uw'
Uw r2 Fw' r' Fw r2 Uw'
They are the bottom red and green wing edge and bottom red and orange wing edge.
[5] Resuming from the end of step 3,
Fw' Uw r' Uw' Fw Uw r Uw'
Uw r2 Fw'
We locate the wings which we know are ultimately swapped in the 2-cycle. They are both in the top layer (in slice r of course). Notice that the red and green wing edge is in the 1x2x2 block which is incomplete. This is the wing edge which we can substitute with another to get an alternate 2-cycle. Any wing edge is permitted except those in slice r. In addition, any moves we do must not affect any piece in slice r except for the red and green wing edge piece in slice r.
Without even worrying about setting up the centers (so that they will not be discolored at all), let’s just grab the green and white wing edge in the top of slice L and carry out the conjugation of the quarter turn.
Fw' Uw r' Uw' Fw Uw r Uw'
[Uw r2 Fw' F' L F: r']
No magic at all. Clearly we can see exactly which wing edges are going to swap and we are no longer dependent on shifting or applying outer conjugates as when forming symmetrical algorithms (although these algorithms are longer, we have total freedom of which wings we swap as well as which centers we swap).
[6] Resuming to the end of step 3, we know that the quarter turn must be r' (see step 2 if you have forgotten) and so the colors of the center pieces of the green and orange 1x2x2 block in the top back quadrant of slice r are the colors we must have in the 1x2x2 block in the top front 1x2x2 block in slice r. REASON: r' brings the top back block to the top front block (I first explained this in Part 4 of the “Holy Grail” Edge Flip algorithm derivation—explanation on interior setup moves). Luckily we already have an orange X-center piece in the top front 1x2x2 block in the exact location the orange X-center pieces from the top back 1x2x2 block will be once the quarter turn r' is executed.
Thus we only need to substitute a green X-center piece in for the red center piece in the top front 1x2x2 block because that’s the color of the other X-center piece in the top back 1x2x2 block. We can only grab the green X-center piece in slices u and L, or the X-center piece in slices l and D. The two green X-center pieces in slice r CANNOT BE TOUCHED. The only restriction you have is to not tough slice r, or if you do, you restore it back EXACTLY (supercube speaking) before you close the conjugation of the quarter turn. The closest green X-center piece (move wise) is the one in slices u and L. So we can use f' L f to do that.
Fw' Uw r' Uw' Fw Uw r Uw'
Uw r2 Fw' f' L f
Closing the conjugation of the quarter turn, r', we have the same original 2-cycle as seen in Step 4, only no centers are discolored anymore (which should not be a shocker now).
Fw' Uw r' Uw' Fw Uw r Uw'
[Uw r2 Fw' f' L f: r']
[7] To make an edge flip algorithm with this which does not color center pieces either, we may again start with:
Fw' Uw r' Uw' Fw Uw r Uw'
Uw r2 Fw'
In Step 5, we found out that the green and red wing edge in the top front is the only wing edge which can be substituted to make a 2-cycle. We also learned that the wing edge that that wing edge swaps with is the orange and green wing edge in the top back portion of slice r. Therefore, we need to substitute the green and red wing edge in the top front with the green and orange wing edge which is in slices u, B, and l. At the same time, in step 6 we found out that the X-center piece we need to substitute for the red X-center piece in the top front 1x2x2 block is the green one in slices L, u and b.
Keeping both of these things in mind, the following moves will satisfy both demands, as well as not affecting any pieces in slice r except for those two slots we are substituting for.
Fw' Uw r' Uw' Fw Uw r Uw'
[Uw r2 Fw' f' U L U' f: r']
= Fw' Uw r' Uw' Fw Uw r Uw' Uw r2 Fw' f' U L U' f r' f' U L' U' f Fw r2 Uw'
= Fw' Uw r' Uw' Fw Uw r' Fw' f' U L U' f r' f' U L' U' f Fw r2 Uw' (23,22)
Since there is so much freedom (the only restriction is, again, to not affect slice r—or whichever inner layer slice the quarter turn needs to be applied to based on your choice of the commutator and beginning conjugation moves), there are multiple paths which can be taken. Here are some more examples of 2-cycles (and higher cycles) I made choosing to start with a three block cycler and then conjugate the quarter turn simultaneously as I moved all three blocks from the three block cycler into the same slice.
Single Edge Flip 2-Cycle Case
[Fw', Uw r' Uw'] [Uw r2 Fw' U L u l Uw': r'] (23,22)
[Fw', Uw r' Uw'] [Uw r2 Fw' f' L Fw R' F': r'] (23,22)
[r', Uw Lw Uw'] [Uw Lw Uw' r Fw2 Lw' Fw F R' F': r'] (25,22)
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw F R' F': r'] (25, 22)
[Lw, Uw' r Uw] [Uw' r2 Fw2 L' Fw' F' R F: r] (25,22)
[Dw r' Dw', Lw] [Lw Uw f' Uw F' u Bw' Dw' f' L' f: l'] (25,24)
[Uw Lw Uw', r'] [r' Uw Bw Uw' Bw L Bw B R' B': r'] (25,25)
[Uw Lw' Uw', r'] [r Fw' Uw Lw' L' Uw' F R' F': r'] (27,26)
[Uw Lw' Uw', r'] [r Fw' L' Uw l' F' L' F Uw': r'] (27,26)
[Fw', Uw r' Uw'] [Uw r2 Fw2 U' L U Fw U R U': r'] (29,26)
[Uw Lw' Uw', r'] [r Fw' U l' Dw F Dw' U2 R U: r'] (31,28)
Directly Opposite 2-Cycle Case
[Fw', Uw r' Uw'] [Uw r2 Fw' f' L f: r'] (19,18)
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw F' L F: r'] (23,20)
[r', Uw Lw Uw'] [Uw Lw Uw' r Fw2 Lw' Fw U' R U: r'] (25,22)
Diagonal 2-Cycle Case
[l', Uw Rw' Uw'] [Uw Rw' Fw Uw B' l2 B: r'] (21,19)
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw F R2 F': r'] (27,22)
Adjacent 2-Cycle Case 1
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw U' R2 U: r'] (27,22)
Adjacent 2-Cycle Case 2
[Fw', Uw r' Uw'] [Uw r2 Fw2 U L U' Fw: r'] (23,20)
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw U' R U: r'] (25,22)
Adjacent 2-cycle Case 3
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw: r'] (19,16)
[Lw, Uw' r Uw] [Uw' r2 Fw2 L' Fw': r] (19,16)
[r', Uw Lw Uw'] [Uw Lw Uw' r Fw2 Lw' Fw: r'] (19,16)
[Uw Lw Uw', r'] [r' Uw Bw Uw' Bw L Bw: r'] (19,19)
Adjacent Double Parity 4-Cycle Case
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw R U R' U' F R' F': r'] (33,30)
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw R F U R' U' R' F': r'] (33,30)
Adjacent Checkerboard Case 2
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw F R' F' U R U': r'] (31,28)
Miscellaneous 4-cycles
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw B L' B': r'] (25,22)
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw U' L2 U: r'] (27,22)
1 2-cycle and 1 3-cycle
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw B' L B: r'] (23,20)
6-cycle
[Fw', Uw r' Uw'] [Uw r2 Fw2 L Fw R U R' U' B' L B: r'] (33,30)
As you might be able to see, alternate 2-cycles can be formed if you replace the wing edge which you place in the quarter turn inner layer slice with the rest of the pieces which was NOT involved with the three block cycler to begin with. The higher cycles (4-cycles, 6-cycles, etc.) and disjoint cycles (1 2-cycle and 1 3-cycle, etc.) can be made by replacing any of the other three wings in the quarter turn inner layer slice.
Although it might be obvious, clearly the only case which this method generates a pretty brief algorithm for is the adjacent 2-cycle case #3, being only one half turn move more than the current minimum of 15 btm (which I established a few months ago with my method of symmetrical wide turn based 2-cycle algorithms).